<span>Last choice on the list:
Object A has a net charge of 0 because the positive and negative
charges are balanced.
Object B has a net charge of –2 because there is an imbalance of
charged particles (2 more negative electrons than positive protons).</span>
Answer:
The the speed of the car is 26.91 m/s.
Explanation:
Given that,
distance d = 88 m
Kinetic friction = 0.42
We need to calculate the the speed of the car
Using the work-energy principle
work done = change in kinetic energy
Put the value into the formula
Hence, The the speed of the car is 26.91 m/s.
Answer:
The launching point is at a distance D = 962.2m and H = 39.2m
Explanation:
It would have been easier with the drawing. This problem is a projectile launching exercise, as they give us data after the window passes and the wall collides, let's calculate with this data the speeds at the point of contact with the window.
X axis
x = Vox t
t = x / vox
t = 7.1 / 340
t = 2.09 10-2 s
In this same time the height of the window fell
Y = Voy t - ½ g t²
Let's calculate the initial vertical speed, this speed is in the window
Voy = (Y + ½ g t²) / t
Voy = [0.6 + ½ 9.8 (2.09 10⁻²)²] /2.09 10⁻² = 0.579 / 0.0209
Voy = 27.7 m / s
We already have the speed at the point of contact with the window. Now let's calculate the distance (D) and height (H) to the launch point, for this we calculate the time it takes to get from the launch point to the window; at this point the vertical speed is Vy2 = 27.7 m / s
Vy = Voy - gt₂
Vy = 0 -g t₂
t₂ = Vy / g
t₂ = 27.7 / 9.8
t₂ = 2.83 s
This is the time it also takes to travel the horizontal and vertical distance
X = Vox t₂
D = 340 2.83
D = 962.2 m
Y = Voy₂– ½ g t₂²
Y = 0 - ½ g t2
H = Y = - ½ 9.8 2.83 2
H = 39.2 m
The launching point is at a distance D = 962.2m and H = 39.2m
In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.
The equation that defines the linear moment is given by
where,
m=Total mass
Mass of Object
Velocity before throwing
Final Velocity
Velocity of Object
Our values are:
Solving to find the final speed, after throwing the object we have
We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.
That way during each section the equations should be modified depending on the previous one, let's start:
A) 
B) 
C) 
Therefore the final velocity of astronaut is 3.63m/s
